Question

Determine whether the given functions form a fundamental set of solutions for the linear system. $$ \mathbf{y}^{\prime}=\left[\begin{array}{rrr} 1 & -2 & 1 \\ 0 & -1 & 1 \\ 0 & 0 & 0 \end{array}\right] \mathbf{y}, \quad \mathbf{y}_{1}(t)=\left[\begin{array}{c} e \\ 0 \\ 0 \end{array}\right], \quad \mathbf{y}_{2}(t)=\left[\begin{array}{c} e \\ e^{-t} \\ 0 \end{array}\right], \quad \mathbf{y}_{3}(t)=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right] $$

Short answer

Answer: Yes, the given functions form a fundamental set of solutions for the given linear system, since their Wronskian determinant is non-zero for all \(t\).

Step-by-step solution

Write down the given functions and Wronskian determinant

The functions provided are: $$\mathbf{y}_{1}(t)=\left[\begin{array}{c} e \\\ 0 \\\ 0 \end{array}\right], \quad \mathbf{y}_{2}(t)=\left[\begin{array}{c} e \\\ e^{-t} \\\ 0 \end{array}\right], \quad \mathbf{y}_{3}(t)=\left[\begin{array}{l} 1 \\\ 1 \\\ 1 \end{array}\right]$$ We'll check the linear independence of these functions by calculating their Wronskian determinant: $$ W(\mathbf{y}_{1}(t),\mathbf{y}_{2}(t),\mathbf{y}_{3}(t)) = \begin{vmatrix} e & e & 1 \\ 0 & e^{-t} & 1 \\ 0 & 0 & 1 \end{vmatrix} $$

Calculate the Wronskian determinant

To compute the determinant, we will use the following method: $$ W(\mathbf{y}_{1}(t),\mathbf{y}_{2}(t),\mathbf{y}_{3}(t)) = e\begin{vmatrix} e^{-t} & 1 \\ 0 & 1 \end{vmatrix} - e\begin{vmatrix} 0 & 1 \\ 0 & 1 \end{vmatrix} + 1\begin{vmatrix} 0 & e^{-t} \\ 0 & 0 \end{vmatrix} $$ Now, calculate the remaining 2x2 determinants: $$ W(\mathbf{y}_{1}(t),\mathbf{y}_{2}(t),\mathbf{y}_{3}(t)) = e(e^{-t}\cdot 1 - 0\cdot 1) - e(0\cdot 1 - 0\cdot 1) + 1(0\cdot 0 - 0\cdot e^{-t}) = ee^{-t} $$

Determine if the Wronskian determinant is non-zero

We calculated the Wronskian determinant to be: $$ W(\mathbf{y}_{1}(t),\mathbf{y}_{2}(t),\mathbf{y}_{3}(t)) = ee^{-t} $$ To check for linear independence, the Wronskian determinant should be non-zero. Since \(e\) and \(e^{-t}\) are positive values for all \(t\), then their product will be non-zero as well. Therefore, $$W(\mathbf{y}_{1}(t),\mathbf{y}_{2}(t),\mathbf{y}_{3}(t)) \neq 0$$ for all \(t\).

Conclusion

The given functions, \(\mathbf{y}_{1}(t)\), \(\mathbf{y}_{2}(t)\), and \(\mathbf{y}_{3}(t)\), form a fundamental set of solutions for the given linear system, since their Wronskian determinant is non-zero for all \(t\).

Key concepts

Linear Independence

Linear independence is a key concept in understanding the behavior of functions or vectors within a mathematical system. In simple terms, a set of functions is said to be linearly independent if no function can be expressed as a combination of the others. This is crucial in differential equations and linear systems, as it ensures that the solutions span the solution space effectively.

For the given problem, the functions \( \mathbf{y}_{1}(t), \mathbf{y}_{2}(t), \mathbf{y}_{3}(t) \) must be checked for linear independence. To achieve this, we calculate the Wronskian determinant. If this determinant is non-zero, the functions are linearly independent, showing that they contribute uniquely to the solution space.

Understanding linear independence helps identify if a fundamental set of solutions appropriately describes all possible solutions to a differential equation. Thus, checking for linear independence is a necessary step in confirming that the functions present the complete picture of the solutions.

Wronskian Determinant

The Wronskian determinant is a powerful tool used in determining the linear independence of a set of functions. Named after Jozef Hoene-Wronski, this determinant provides insight into whether a set of solutions is unique or if each component function adds a unique element to the overall solution.

In the context of the initial problem, the Wronskian is calculated as the determinant of a matrix composed of the functions \( \mathbf{y}_{1}(t), \mathbf{y}_{2}(t), \mathbf{y}_{3}(t) \). A key point is that if this Wronskian determinant is non-zero over a range of interest, the functions are linearly independent.

To calculate the Wronskian, you form a matrix with the functions as columns. In our example, the determinant calculated as \( W(\mathbf{y}_{1}(t),\mathbf{y}_{2}(t),\mathbf{y}_{3}(t)) = ee^{-t} \) is non-zero, confirming the independence of the functions.

Linear Systems

Linear systems refer to systems of equations where the variables appear to the first power and interactions between variables are simple additions or subtractions. In mathematics, linear systems play a vital role in modeling and solving real-world problems. They are characterized by their predictability and ease of calculation, often represented by matrices and vectors.

In the problem at hand, our linear system is expressed by the matrix equation:
\( \mathbf{y}^{\prime}=[A]\mathbf{y} \), where \( [A] \) is the coefficient matrix. Solving such a system generally involves identifying a set of solutions - named a fundamental set.

To form a complete solution for such systems, linearly independent functions must be selected. Linear independence ensures that the solutions are robust and can be combined to form any possible solution of the system, emphasizing why it is critical to verify this in the previous steps.

Differential Equations

Differential equations are equations that involve unknown functions and their derivatives. They are exceedingly important in describing various phenomena such as motion, growth, decay, and much more. The main task is to determine the function or functions that satisfy the given differential equation.

In our original exercise, the differential equation is part of a linear system, expressed in terms of matrix notation. Differential equations within linear systems can be efficiently handled using matrices and determinants, leveraging the properties of linear algebra.

When given a set of functions as potential solutions, it is important to check their linear independence using the Wronskian. This helps in ascertaining that the set forms a fundamental set of solutions for the linear differential system, ensuring that any valid solution to the differential equation can be constructed from them.